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Intercepts on the Axes made by a Circle

We will learn how to find the intercepts on the axes made by a circle.

The lengths of intercepts made by the circle x2 + y2 + 2gx + 2fy + c = 0 with X and Y axes are 2√g2βˆ’c and 2√f2βˆ’c respectively.

Proof:

Let the given equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 ………. (1)

Clearly, the centre of the circle is c (-g, -f) and the radius = √g2+f2βˆ’c

Let AB be the intercept made by the given circle on x-axe. Since on x-axis, y = 0. Therefore, x-coordinates of the points A and B are the roots of the equation x2 + 2gx + c = 0.

Let x1 and x2 be the x-coordinates of the points A and B respectively. Then, x1 and x2 also the roots of the equation x2 + 2gx + c = 0.

Therefore, x1 + x2 = - 2g and x1x2 = c

Clearly the intercept on x-axis = AB

                                          = x2 - x1 = √(x2βˆ’x1)2

                                          = √(x2+x1)2βˆ’4x1x2

                                          = √4g2βˆ’4c

                                          = 2√g2βˆ’c

Therefore, the intercept made by the circle (1) on the x-axis = 2√g2βˆ’c

Again,

Let DE be the intercept made by the given circle on y-axe. Since on y-axis, x = 0. Therefore, y-coordinates of the points D and E are the roots of the equation y2 + 2fy + c = 0.

Let y1 and y2 be the x-coordinates of the points D and E respectively. Then, y1 and y2 also the roots of the equation y2 + 2fy + c = 0

Therefore, y1 + y2 = - 2f and y1y2 = c

Clearly the intercept on y-axis = DE

                                          = y2 - y1 = √(y2βˆ’y1)2

                                          = √(y2+y1)2–4y1y2

                                          = √4f2βˆ’4c

                                          = 2√f2βˆ’c

Therefore, the intercept made by the circle (1) on the y-axis = 2√f2βˆ’c

Solved examples to find the intercepts made by a given circle on the co-ordinate axes:

1. Find the length of the x-intercept and y-intercept made by the circle x2 + y2 - 4x -6y - 5 = 0 with the co-ordinate axes.

Solution:

Given equation of the circle is x2 + y2 - 4x -6y - 5 = 0.

Now comparing the given equation with the general equation of the circle x2 + y2 + 2gx + 2fy + c = 0, we get g = -2 and f = -3 and c = -5

Therefore, length of the x-intercept = 2√g2βˆ’c = 2√4βˆ’(βˆ’5) = 2√9 = 6.

The length of the y-intercept = 2√f2βˆ’c = 2√9βˆ’(βˆ’5) = 2√14.


2. Find the equation of a circle which touches the y-axis at a distance -3 from the origin and cuts an intercept of 8 units with the positive direction of x-axis.

Solution:

Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 …………….. (i)

According to the problem, the equation (i) touches the y-axis

Therefore, c = f2 ………………… (ii)

Again, the point (0, -3) lies on the circle (i).

Therefore, putting the value of x = 0 and y = -3 in (i) we get,

9 - 6f + c = 0 …………………… (iii)

From (ii) and (iii), we get 9 - 6f + f2 = 0 β‡’ (f - 3)2 = 0 β‡’ f - 3 = 0 β‡’ f = 3

Now putting f = 3 in (i) we get, c = 9

Again, according to the problem the equation of the circle (i) cuts an intercept of 8 units with the positive direction of x-axis.

Therefore,

2√g2βˆ’c = 8

β‡’ 2√g2βˆ’9 = 8

β‡’ βˆšg2βˆ’9 = 4

β‡’ g2 - 9 = 16, [Squaring both sides]

β‡’ g2 = 16 + 9

β‡’ g2 = 25

β‡’ g = Β±5.

Hence, the required equation of the circle is x^2 + y^2 Β± 10x + 6y + 9 = 0.

● The Circle




11 and 12 Grade Math 

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